1. ## Rings and Ideals

Need some help...

Let $\displaystyle \mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1)Show that $\displaystyle <3>=<3;1+\sqrt{-5}><3;1-\sqrt{-5}>$
($\displaystyle <a>$ means the ideal generated by $\displaystyle a$)

2)Prove that the ideal on the right is prime.

Thank you folks!

2. Originally Posted by Inti

Let $\displaystyle \mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1) Show that $\displaystyle <3>=<3,1+\sqrt{-5}><3,1-\sqrt{-5}>$
Hint: we have $\displaystyle 3=3^2 - (1+\sqrt{-5})(1-\sqrt{-5}).$ thus: $\displaystyle <3> \subseteq <3,1+\sqrt{-5}><3,1-\sqrt{-5}>.$ for the other side of the inclusion, note that $\displaystyle (1+\sqrt{-5})(1-\sqrt{-5})=6 \in <3>.$

2) Prove that the ideals on the right is prime.
Hint: define the map $\displaystyle f: \mathbb{Z}[\sqrt{-5}] \longrightarrow \mathbb{Z}_3$ by $\displaystyle f(a+b\sqrt{-5})=a-b.$ show that $\displaystyle f$ is a well-defined surjective ring homomorphism and: $\displaystyle \ker f = <3, 1+\sqrt{-5}>.$

3. thereīs one thing thatīs not clear to me:
Iīve just proved that $\displaystyle \ker f = <3,1+\sqrt{-5}>$, which means that $\displaystyle \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\displaystyle \mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?
As far as Iīm concerned, the quotient of a ring by a prime ideal is a domain.
Iīm kind of confused with that...
But the help was very helpful

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??

Thanks!

4. Originally Posted by Inti

thereīs one thing thatīs not clear to me:
Iīve just proved that $\displaystyle \ker f = <3,1+\sqrt{-5}>$, which means that $\displaystyle \mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\displaystyle \mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?
that means the ideal is maximal and we know that every maximal ideal is prime.

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??
no. if $\displaystyle I$ and $\displaystyle J$ are two ideals, then $\displaystyle IJ=\{\sum_{k=1}^na_kb_k: \ n \in \mathbb{N}, \ a_k \in I, b_k \in J \}.$ however, for example, if you want to prove that $\displaystyle IJ \subseteq K,$ where $\displaystyle K$ is an ideal, then you only need to prove that

$\displaystyle ab \in K$ for all $\displaystyle a \in I, b \in J.$